Problem: $\log_{4}16 = {?}$
Solution: If $\log_{b}x=y$ , then $b^y=x$ First, try to write $16$ , the number we are taking the logarithm of, as a power of $4$ , the base of the logarithm. $16$ can be expressed as $4\times4$ $16$ can be expressed as $4^2$ $4^2=16$, so $\log_{4}16=2$.